Masha and Grisha like studying sets of positive integers.

One day Grisha has written a set A containing n different integers ai on a blackboard. Now he asks Masha to create a set B containing n different integers bj such that all n2 integers that can be obtained by summing up ai and bj for all possible pairs of i and j are different.

Both Masha and Grisha don’t like big numbers, so all numbers in A are from 1 to 106, and all numbers in B must also be in the same range.

Help Masha to create the set B that satisfies Grisha’s requirement.

Input
Input data contains multiple test cases. The first line contains an integer t — the number of test cases (1 ≤ t ≤ 100).

Each test case is described in the following way: the first line of the description contains one integer n — the number of elements in A (1 ≤ n ≤ 100).

The second line contains n integers ai — the elements of A (1 ≤ ai ≤ 106).

Output
For each test first print the answer:

NO, if Masha’s task is impossible to solve, there is no way to create the required set B.
YES, if there is the way to create the required set. In this case the second line must contain n different positive integers bj — elements of B (1 ≤ bj ≤ 106). If there are several possible sets, output any of them.
Example
Input
3
3
1 10 100
1
1
2
2 4
Output
YES
1 2 3
YES
1
YES
1 2

給你一個集合A，讓你構造一個集合B，使得A中任何一個元素+B中任何一個元素的所有能產生的和都不相同。轉化一下 ai+bj！=ak+bm; ai-ak!=bj-bm;

找出所有的差值存下，然後掃一遍所有的數，如果能成立就放進去，一開始 b[0]=1;

#include <bits/stdc++.h>
using namespace std;

int vis[1000010];
int b[330];
int a[330];

const int BUF=30000000;  
char Buf[BUF],*buf=Buf; 
inline void read(int&a){for(a=0;*buf<48;buf++);while(*buf>47)a=a*10+*buf++-48;}


int main()
{
    int t;
    fread(Buf,1,BUF,stdin);//重點  
    read(t);
    while(t--)
    {
        int n;
        read(n);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            read(a[i]);
        int cnt=1;
        b[0]=1;
        if(n==1)
        {
            puts("YES");
            printf("%d\n",b[0] );
            continue;
        }
        sort(a+1,a+n+1);
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
                vis[a[j]-a[i]]=1;
        for(int i=2;i<=1000000;i++)
        {
            int f=0;
            for(int j=0;j<cnt;j++)
            {
                if(vis[i-b[j]])
                {
                    f=1;
                    break;
                }
            }
            if(!f)
                b[cnt++]=i;
            if(cnt>=n)
                break;
        }
        if(cnt>=n)
        {
            puts("YES");
            for(int i=0;i<cnt;i++)
                printf("%d ",b[i] );
            printf("\n");
        }
        else puts("NO");
    }
}